What is the extraneous solution to these equations? $\dfrac{x^2 + 1}{x + 5} = \dfrac{-8x - 14}{x + 5}$
Solution: Multiply both sides by $x + 5$ $ \dfrac{x^2 + 1}{x + 5} (x + 5) = \dfrac{-8x - 14}{x + 5} (x + 5)$ $ x^2 + 1 = -8x - 14$ Subtract $-8x - 14$ from both sides: $ x^2 + 1 - (-8x - 14) = -8x - 14 - (-8x - 14)$ $ x^2 + 1 + 8x + 14 = 0$ $ x^2 + 15 + 8x = 0$ Factor the expression: $ (x + 5)(x + 3) = 0$ Therefore $x = -5$ or $x = -3$ At $x = -5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -5$, it is an extraneous solution.